题意：给你n个模式串，再给你m个主串，问你每个主串中有多少模式串，并输出是哪些。注意一下，这里给的字符范围是可见字符0~127，所以要开130左右。

思路：用字典树开的时候储存编号，匹配完成后set记录保证不重复和排序。

代码：

#include<cstdio>
#include<vector>
#include<set>
#include<queue>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#define ll long long
const int maxn = 5000000+5;
const int maxm = 100000+5;
const int MOD = 1e7;
const int INF = 0x3f3f3f3f;
const int kind = 130;
const char baset = 0;
using namespace std;
struct Trie{
    Trie *next[kind];
    Trie *fail; //失配值
    int sum;    //以此为单词结尾的个数
    int id;
    Trie(){
        sum = 0;
        memset(next,NULL,sizeof(next));
        fail = NULL;
        id = 0;
    }
};
Trie *root;
queue<Trie *> Q;
int head,tail;
set<int> record;
void Insert(char *s,int id){
    Trie *p = root;
    for(int i = 0;s[i];i++){
        int x = s[i] - baset;
        if(p ->next[x] == NULL){
            p ->next[x] = new Trie();
        }
        p = p ->next[x];
    }
    p ->sum++;
    p ->id = id;
}
void buildFail(){   //计算失配值
    while(!Q.empty()) Q.pop();
    Q.push(root);
    Trie *p,*temp;
    while(!Q.empty()){
        temp = Q.front();
        Q.pop();
        for(int i = 0;i < kind;i++){
            if(temp ->next[i]){
                if(temp == root){   //父节点为root，fail为root
                    temp ->next[i] ->fail = root;
                }
                else{
                    p = temp ->fail;    //查看父节点的fail
                    while(p){
                        if(p ->next[i]){
                            temp ->next[i] ->fail = p ->next[i];
                            break;
                        }
                        p = p ->fail;
                    }
                    if(p == NULL) temp ->next[i] ->fail = root;
                }
                Q.push(temp ->next[i]);
            }
        }
    }
}
void ac_automation(char *ch){
    //p为模式串指针
    Trie *p = root;
    int len = strlen(ch);
    for(int i = 0;i < len;i++){
        int x = ch[i] - baset;
        while(!p ->next[x] && p != root)
            p = p ->fail;
        p = p ->next[x];    //找到后p指针指向该结点
        if(!p) p = root;    //若指针返回为空，则没有找到与之匹配的字符
        Trie *temp = p;
        while(temp != root){
            if(temp ->id > 0)
                record.insert(temp ->id);
            temp = temp ->fail;
        }
    }
}
char ch[210];
char s[10005];
int main(){
    int n,m,x;
    root = new Trie();
    scanf("%d",&n);
    for(int i = 1;i <= n;i++){
        scanf("%s",ch);
        Insert(ch,i);
    }
    buildFail();
    int tot = 0;
    scanf("%d",&m);
    for(int i = 1;i <= m;i++){
        record.clear();
        scanf("%s",s);
        ac_automation(s);
        if(record.size()){
            tot++;
            printf("web %d:",i);
            int End = record.size();
            for(int j = 1;j <= End;j++){
                int id = *record.begin();
                printf(" %d",id);
                record.erase(id);
            }
            printf("\n");
        }
    }
    printf("total: %d\n",tot);
    return 0;
}